Modeling delayed feedback using logistic regression - Cross Validated most recent 30 from stats.stackexchange.com 2019-07-23T07:55:59Z https://stats.stackexchange.com/feeds/question/289345 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/289345 4 Modeling delayed feedback using logistic regression decay https://stats.stackexchange.com/users/19651 2017-07-07T14:22:15Z 2019-06-07T14:01:30Z <p>Suppose we are trying to model the probability of a user clicking on an ad using logistic regression. We will receive only the positive feedback so, we define $Y = 1$ when success was observed, $Y=0$ otherwise.</p> <p>We define the probability of click for a set of features $X$ as $$P(y_i=1|X=x_i) = \frac{1}{1+e^{-wx_i}}$$</p> <p>Now suppose that you should predict the probability of click in real-time, one-by-one when the ad is displayed but the feedback will be delayed. And, in the same way, when you retrieve the data for training you will have actions with no feedback and therefore marked as $Y=0$ but you could receive positive feedback hours later and the label will change to $Y=1$.</p> <p>Note that the data is non stationary because new features values (or instances) can appear within minutes, so training with data old enough to make sure that you received all the possible positive feedback is not an option.</p> <p>Here is an example showing the cumulative amount of clicks received per hours. As you can see we have received almost 25% of clicks in the first hour and the 85% in the 10th.</p> <p><a href="https://i.stack.imgur.com/O7tbt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7tbt.png" alt="Feedback decay"></a></p> <p>The image is showing the actual decay (in red) and the exponential decay we are using to model it (in yellow). Following:</p> <p>$$N(t) = N_0 · e^{-\lambda t}\\ \lambda=-ln(0.15)/C\\$$</p> <p>where $N_0$ is the initial point and $C$ is elapsed hours to 85% (that's why $\lambda$ is computed using 0.15)</p> <p>In this paper <a href="https://pdfs.semanticscholar.org/9eed/333dcaade2c1dd32590eff6b0c0e670a5674.pdf" rel="nofollow noreferrer">Modeling Delayed Feedback in Display Advertising</a> they introduced the delay into the model itself but I am trying to use the exponential decay to model the output variable of the logistic regression for simplicity (I though it'd be easier to change labels than rewrite the optimizer). So, instead of</p> <p>$$y = \left\{\begin{matrix} 1 &amp; \text{success observed}\\ 0 &amp; \text{otherwise} \end{matrix}\right.$$</p> <p>I am trying to train the model using</p> <p>$$y = \left\{\begin{matrix} 1 &amp; \text{success observed}\\ N_0 · e^{-\lambda t} &amp; \text{otherwise} \end{matrix}\right.$$</p> <p>and setting $N_0$ as the average successful rate.</p> <p>The problem is that I haven't seen any place where this method was used and I don't know if I am doing something terribly wrong.</p> <ul> <li>Is this approach valid? </li> <li>Any suggestion or different approach to introduce the delayed feedback in a logistic regression model?</li> <li>Logistic regression is good fitting binomial distributions but here I am using <em>soft</em> labels ($y \in (0, 1)$ instead of $y \in \{0, 1\}$). Is this approach valid or logistic regression isn't gonna work well?</li> </ul> https://stats.stackexchange.com/questions/289345/-/291181#291181 0 Answer by user1151446 for Modeling delayed feedback using logistic regression user1151446 https://stats.stackexchange.com/users/76532 2017-07-12T14:57:19Z 2017-07-12T14:57:19Z <p>your problem reminds me a lot the delayed feedback in conversion modeling published by Olivier Chapelle few years ago: <a href="http://dl.acm.org/citation.cfm?doid=2623330.2623634" rel="nofollow noreferrer">http://dl.acm.org/citation.cfm?doid=2623330.2623634</a> also available here: <a href="http://olivier.chapelle.cc/pub.html" rel="nofollow noreferrer">http://olivier.chapelle.cc/pub.html</a></p> <p>The only difference is that you work with clicks and not conversions but the idea is the same. There are also simple baselines in his paper that could work for you. Hope it helps.</p>